How do you find the magnitude and direction angle of the vector $v=-5i+4j$ ?

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Answer 1 · 2017-08-03T13:50:48

$|vec v|=sqrt(41)*unit$
$theta = pi/2+tan^-1(5/4)$
$~~2.46685 rad~~141.34^@~~141^@ 20' 24.69''$

$|vec v|=sqrt(x^2+y^2)=sqrt((-5)^2+4^2)=sqrt(25+16)=sqrt(41)*unit$

Now it is clear that the head of the vector lies in the second quadrant, and we can use a little trigonometry to find out the angle.

$theta = 90^@+tan^-1(5/4)$

How do you find the magnitude and direction angle of the vector v=-5i+4jv=-5i+4j?